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PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 67. Number 2, December 1977 A TREE ARGUMENTIN INFINITARYMODEL THEORY1 V. HARNIK AND M. MAKKAI Abstract. A tree argument is used to show that any counterexample to Vaught's conjecture must have an uncountable model. A similar argument replaces the use of forcing by Burgess in a theorem on 2j equivalence relations. A.formula or a sentence is one of Lu u. A sentence </>is a counterexample to Vaught's conjecture or, simply, a counterexample if it has more than «0 but less than 2"° nonisomorphic countable models. A sentence is large if it has more than N0nonisomorphic countable models. A large sentence is minimal if for every sentence \¡/, either <j> /\ \j/ or <b/\ -| v//is not large. Theorem 1. Every counterexample can be strengthened to a minimal counterexample, i.e., if a is a counterexample, then there is a minimal counterexample 0 such that <í>|= a. The proof of Theorem 1 uses a lemma due to Morley [8], whose formulation depends on the following (stronger than usual) notion of fragment. A set of formulas A is a fragment if it is closed under subformulas, substitutions of terms, finitary logical operations and if it satisfies: whenever <j)EA,V6eA (where6 c A),then \/{3x9: 9 E 9}, V{</> A 8: 9 E S), V ({</>} U @)all belong to A. Lemma 2 [8]. Let A be a countable fragment. If T e àis a finitely consistent set of sentences such that for all valid V® e A tnere is a 9 E ® which belongs to T, then T is consistent. Proof of Theorem 1. Assume that the theorem is false for some a. Then it is easily seen that there is a countable fragment A containing a and such that for every large <bE A s.t. <J> f= a, there is »//E A with both <b/\ \¡/and <f> /\ -\\}/ large. Let {8„}n<u be an enumeration of all valid disjunctions belonging to A. We are going to define a tree Ts, s E <u2 ( = the set of finite sequences of O's and l's) such that for all í: (a) Ts is a finite subset of A, a G Ts and /\Tsis& (b) TsA(0yand TsA<Xy are contradictory; and large sentence; (c) if lh s = i and 5, = V© then there are 9', 9" E 0 s.t. 9' E TsA<oy and Received by the editors April 6, 1977. AMS (A/05) subjectclassifications(1970).Primary 02B25. 'Research partially supported by the National Research Council of Canada. © American Mathematical Society 1978 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 310 V. HARNIK AND M. MAKKAI Once the tree is constructed, consider, for each tj £"2 (= the set of infinite sequences of 0's and l's), the set Tv = Ui^f,: » < w}- By (a), (c) and Lemma 2, each Tv is consistent, hence has a countable model M^ (which, by (a), is a model of a); by (b), if tj t^ tj' then Mn aé M^,. Hence (A/,,: tj Ew2} is a collection of 2"° nonisomorphic countable models of a, a contradiction to the assumption that a is a counterexample. Thus, to conclude the proof, we have to indicate how to get Ts. This is done by induction on the length of s. Take T^y = {a}. Assume that Ts has been defined and lh s = /'. As /\ Ts E A is large and implies a, there is \p E A s.t. both A^A'i' and A^Ai^ are large. If 5, = V ® then any of the uncountably many models of /\TS A \L (/\TS A -j \p) is a model of some A^, AM^A^AiM*)0) s.t. /\Ts/\*/\9' with9 E 0. Thus,thereis 9' e 6 (9" E (A^AnMn is large. Take 7><0>= Ts u The proof of Theorem 1 is now complete. Given a counterexample a, call a formula (#>(x)with free variables x (x a finite sequence) large (with respect to a) if a A 3x<Xx) is large. Call <b(x) minimal (w.r.t. a) if for all >Kx), <j>A<//or<r>A-|,r/is not large. Theorem 1 yields Corollary 3. Every large (w.r.t. a given counterexample a) formula <b(x) can be strengthened to a minimal formula. Proof. Notice that <b(x) is large iff a A «Kc) (a sentence in the larger language L(c) with c a sequence of new constants) is a counterexample and that <Kx) is minimal w.r.t. o iff a A <KC)is a minimal counterexample. Now, the assertion follows from Theorem 1. Now assume that o is a minimal counterexample. For any fragment A containing o, define TA = (<#>:ij>6A and <j> /\o is large}. If A is countable then, by the minimality of a, T¿ is consistent (all but countably many of the countable models of a are models of T¿) and A-complete. A formula ^(x) G A is consistent with TAiff it is large. This observation easily yields Lemma 4. // <b(x) is minimal, then for all countable A, // <f>G A then <j>is complete (in A) with respect to T¿ (cf. the definition on p. 61 o/[4]). Call a fragment A closed if for every large <Kx) G A there is a minimal formula <i>'(x)G A s.t. N <i>'-» <i>.By Corollary 3, every countable fragment can be enlarged to a closed one. If A is countable and closed then, by Lemma 4 and the definition of 7'A, there are no incompletable formulas ^(x) with respect to TA. Hence, rA has a prime model (cf. [4, pp. 61-64, especially Theorem 16]). Moreover, it is easily seen that if A is closed then a formula <j>G A is complete w.r.t. 7¿ iff it is minimal. Hence, each finite sequence of elements in the prime model of TAsatisfies a minimal formula belonging to A. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use infinitary model theory 311 As an application, we can now prove the following result (announced in [2])Theorem 5. If a is a counterexample then it has an uncountable model N. Moreover, N can be so chosen as to satisfy only large sentences; thus, N is not LMUequivalent to any countable structure. Proof. By Theorem 1, we may assume that a is minimal. We define by induction an increasing chain of countable fragments Aa, and of countable structures Ma, a < u, s.t.: (i) a £ Aa, and Aa is closed. (ii) Ma is the prime model of TA . (iii) Ma ¥= Ma+X, Ma <^ Ma+, and Mx = U a<KMa for a limit X. Once this is done, we shall take N = (J o<10Ma. The inductive definition goes as follows: Assume that Aa and Ma are defined. Let Aa+1 D Aa be a closed fragment which contains a Scott sentence </>of Ma (cf. Chapter 2 in [4]). Let Ma+X be the prime model of 7^. As Ma+1(= rAo (c 7a„+1)an<i Ma is a prime model of 7A , Ma can be embedded in Ma+X. Thus, we may take Ma <Aa Ma+X. As the Scott sentence <t>of Ma is obviously not large, <££ rA , hence Ma+X (= -| <b.This shows that Ma sé A/a+1, hence, M„^Afa+1. For a limit X, take AA= Ua<AA„ and Mx = UaexMx- Then AA is obviously closed and every finite sequence of elements of Mx satisfies a minimal, hence by Lemma 4, complete (w.r.t. T¿J formula. It follows (again by Theorem 16 in [4]) that Mx is the prime model of Tà . As said before, we take N = (J a<a Ma to get an uncountable model of a. If N 1= <bthen there is a closed and unbounded set C E ux s.t. Ma ^= <bfor all a E C. By construction, Ma sá Mß whenever a ^ ß; hence [Ma: a E C) is an uncountable collection of nonisomorphic countable models of <f>,showing that <j>is large. This completes the proof of Theorem 5. An equivalent formulation of Theorem 5 says that, for a minimal counterexample a, TAis consistent even for uncountable fragments A, in particular for A = LUiU. The model property that implies that N such that for N constructed in the proof of Theorem 5 has the further each finite sequence of it satisfies a minimal formula. This has Scott height w, (the Scott height of a structure is the first a all finite sequences a of elements of N, N ^= Vx(<i>£(x)-» (f>a+l(x)) where <}>£ is the Lxu formula defined as in Chapter 2 of [4], without the restriction a < ux). Independently, Leo Harrington showed the stronger result that every counterexample has uncountable models of arbitrarily large Scott heights a < u2 (unpublished). The second author of the present paper showed [5] that every counterexample has an uncountable model which is L^-equivalent to a countable one. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 312 V. HARNIK AND M. MAKKAI In a somewhat different context, we wish now to point out that the same tree argument that went into the proof of Theorem 1 can be used to replace Burgess' use in [1] of forcing in the deduction of his result on 2¡ equivalence relations from Silver's theorem on n¡ equivalence relation [10] (cf. also [3]). By an equivalence relation E we mean one defined between countable L-structures which is weaker than isomorphism, i.e., if M ss N then M(E)N. Let L be the (two sorted) language corresponding to the naturally defined disjoint sum M © A of any two L-structures M, N. The equivalence relation E is Borel (resp. 2¡) if for some a G L'a¡a (resp. a(P) G L'(P)U¡U),M(E)N iff M ® N\= a (M 0 A (= 3Po(P)). The following is an observation of Burgess: Claim 6. Every 2j equivalence relation is the intersection of N, Borel equivalence relations. Sketch of a model-theoretic proof. By Vaught [11] (see also [6]), there are L'a u sentences </>„,a < ux, such that on countable structures 3Po(P) is equivalent to A0<Ul</V Set M(Ea)N iff M © N (= <j>a. The claim will follow if we show that Ea is an equivalence relation whenever a is an (admissible) ordinal such that a(P) is &-finite for some admissible set with ords = a. But this is easily established using the existence of Sg-saturated models, as well as the fact that, in any ^saturated structure, <ba(= /\ß<a<Pß) is equivalent to 3Pa(P) (cf. Corollary 7.3 and the proof of 8.1 in [6]; the notion of 2asaturated structure comes from [9]). Using Claim 6 and a forcing argument, Burgess deduced from Silver's aforementioned theorem that any Sj equivalence relation has < x, or 2"° equivalence classes. This result follows from the following. Theorem 7. Assume that N0 < k < 2"°. // E is the intersection of k many Borel equivalence relations then E has < k or 2"° many equivalence classes. Proof. Let E = n a<KEa, each Ea a Borel equivalence relation; as is well known and easily seen, every equivalence class of any Ea is also Borel, i.e. definable by an L,,,, sentence. Assume that E has < 2"° equivalence classes. Then the same is true for each Ea and by (a weakened version of) Silver's theorem, Ea has < «0 equivalence classes. For every equivalence class X of any Ea, a < k, select an LMU sentence defining it; collect all these sentences into a set ¥. The properties of ^ are summed up as follows: 0)1*1< «; (ii) each \p E ^ is E-invariant, i.e., M(= i/<and M(E)N imply that A(= \p; and (iii) ^ distinguishes between the equivalence classes of E; i.e. if -\M(E)N, then for some x¡/E ¥, M |= $ and N \= -| \p. Assume next, for proof by contradiction, that E has > k equivalence classes. For <|>G Luu, let G^ be the set of equivalence classes of E having a nonempty intersection with Mod(<i>).Call o large iff Qa has power > k. A License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use INFINITARY MODEL THEORY 313 simple counting argument shows: Claim 8. For every large a there is \p E ^ s.t. both o A 'I' and o A ~| ^ are large. Proof. Let T be the set of those sentences y which belong to ¥ or are negations of members of ^ s.t. a /\y is not large. Then e0=U{eaAy:yET}u n{(Wyer}. If the claim is false, then the second term of the union is easily seen, by (hi) above, to contain just one element. Thus, by (i), |6„| < k, contradicting the largeness of a. Now, using Claim 8, one constructs precisely as in the proof of Theorem 1, a tree 7;, s E<"2 s.t. for all s: (a) Ts is finite, a E Ts and /\TS is large; (b) there is \p E ^ s.t. \L E TsA<0y and -jt// E rjA<1>,and such that for all r, E"2, r„ - U n<aTMn has a model Mr By (ii) and (b) above, ^M^M^. whenever tj ^ i)'. Thus E has 2"° equivalence classes, a contradiction. We conclude by indicating a most natural example of a relation which is the intersection of K, Borel equivalence relations. If K is the class of countable members of a PCU class, define: M(E)N iff M, N E K or M ss N. Again by [11], K = n„<U|K„ where each K0 is Borel. Define: M(Ea)N iff M, N E K„ or M =a N (where =a means equivalence w.r.t. Lu¡u sentences with quantifier rank < a). Obviously, E = f) a<a Ea and each Ea is Borel. Thus, Theorem 7 implies Morley's result [7] that any PCU a class has < N, or 2"° countable nonisomorphic models. (Actually, Morley's proof contains an argument showing that the particular E0 defined above satisfies Silver's theorem.) Added May 30, 1977. In [12], John Burgess proves a theorem which is more general than our Theorem 7. Burgess' proof is forcing-free as well. We are indebted to the referee for this information. References 1. J. P. Burgess, Infinitary languages and descriptive set theory, Ph. D. Thesis, Univ. of California, Berkeley, 1974. 2. V. Harnik and M. Makkai, Some remarks on Vaught's conjecture, J. Symbolic Logic 40 (1975),300-301(abstract). 3. L. Harrington, A powerlessproof of a theorem of Silver (manuscript). 4. H. J. Keisler, Model theoryfor infinitary logic, North-Holland, Amsterdam, 1971. 5. M. Makkai, An "admissible" generalization of a theorem on countable 2} sets of reals with applications,Ann. of Math. Logic 11 (1977), 1-30. 6. _, Admissible sets and infinitary logic, Handbook of Logic (J. K. Barwise, editor), North-Holland, Amsterdam, 1977. 7. M. Morley, The numberof countablemodels,J. Symbolic Logic 35 (1970), 14-18. 8. _, Applications of topology to L„|U, Proc. Sympos. Pure Math., vol. 25, Amer. Math. Soc., Providence,R. I., 1973,pp. 233-240. 9. J.-P. Ressayre, Models with compactness properties with respect to logics on admissible sets, Ann. of Math. Logic 11 (1977),31-55. 10. J. Silver, Any Tl\ equivalence relation over 2" has either 2"° or < H0 equivalence classes (manuscript). License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 314 V. HARNIK AND M. MAKKAI U.R. Vaught, Descriptive set theory in La¡u, Lecture Notes in Math., vol. 337, Springer-Verlag, Berlinand New York, 1973,pp. 574-598. 12. J. P. Burgess, Equivalences generated by families of Borel sets, Proc. Amer. Math. Soc. (to appear). Department of Mathematics, University of Haifa, Haifa, Israel Department of Mathematics, McGill University, Montreal, Quebec, Canada License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use